1.

An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0 s apart is 30°, what is the speed of the aircraft ? (tan = 15° = 0.2679)

Answer» <html><body><p></p>Solution :Let O be the <a href="https://interviewquestions.tuteehub.com/tag/ground-1013213" style="font-weight:bold;" target="_blank" title="Click to know more about GROUND">GROUND</a> observation point and A, B, C be the <a href="https://interviewquestions.tuteehub.com/tag/positions-1159905" style="font-weight:bold;" target="_blank" title="Click to know more about POSITIONS">POSITIONS</a> of the aircarft at t = <a href="https://interviewquestions.tuteehub.com/tag/0s-256530" style="font-weight:bold;" target="_blank" title="Click to know more about 0S">0S</a>, t = 5s and t = 10 s respectively, <br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_NEO_CAO_PHY_XI_V01_MP1_C04_SLV_022_S01.png" width="80%"/> <br/> Clearly, `tan15^(@) = (AB)/(OB)` or `AB= OB tan 15^(@)` <br/> <a href="https://interviewquestions.tuteehub.com/tag/thus-2307358" style="font-weight:bold;" target="_blank" title="Click to know more about THUS">THUS</a>, distance travelled by the <a href="https://interviewquestions.tuteehub.com/tag/aircraft-3321" style="font-weight:bold;" target="_blank" title="Click to know more about AIRCRAFT">AIRCRAFT</a> in 10s, <br/> i.e., `s = AC = 2AB = 2(OB tan 15^(@))` <br/> `= 2 xx 3400 xx 0.2679 = 1822 m` <br/> Speed of the aircraft `= (s)/(t) = (1822m)/(10s) = 182.2 m//s`</body></html>


Discussion

No Comment Found

Related InterviewSolutions